Ask question

Ask question

All categories

3 years ago

7

What happens to the temperature of a gas when it is compressed? a the temperature does not change. b the temperature increases.

c the temperature becomes unpredictable. d the temperature decreases?

2 answers:

kupik [55]3 years ago

8 0

Answer:

<h2>Increases.</h2>

Explanation:

Because the volume is reduced, particles are accelerated and that increases the temperature, due to the energy inside the system.

nasty-shy [4]3 years ago

5 0

I think the temperature increases

You might be interested in

It’s 11:53pm right now and I have just this assignment left to do for this class, I need this assignment done so I don’t fail th

Vlada [557]

Answer:

Explanation:

1 matter

2 the chain / air pressure

3 gravity

4 in the cup

5 inertia

6 pushed or pulled

7 rest

8 motion

9 mass

10 reaction

11 b

12 a

13 grams

14 listen to the song

7 0

3 years ago

Read 2 more answers

What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me

alexgriva [62]

Concept: The magnification of spherical mirror can be defined by two ways.

(i) In terms of the height of the object and image.

The magnification of the spherical mirror is defined as the ratio of the height of the image' $h_{i}$ ' to the height of the object ' $h_{o}$ '. It is denoted by letter 'm'.

Mathematically, it can be written as

$m= \frac{h_{i}}{h_{o}} ------------(1)$

(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance' $d_{i}$ ' to the object distance ' $d_{o}$ '.

Mathematically, it can be written as

$m= - \frac{d_{i}}{d_{o}} ------------(2)$

Now, from equation (1) and (2) we have,

$m = \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} -----------(3)$

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

Object distance $(d_{o}) = + 8 mm$ .

Image distance $(d_{i}) = + 16 mm$

Object's height $(h_{o}) = + 4 mm$

Image's height $(h_{i}) =?$

Now, apply equation (3)

$\frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}}$

$Or, \frac{h_{i}}{4 mm} = - \frac{+16 mm}{+8 mm}$

Or, hi = - 8 mm

Here; negative sign means, the image will be inverted.

The image's height will be 8 mm.

4 0

3 years ago

In which situation is maximum work considered to be done by a force

Katena32 [7]

Answer:

Work done is maximum when the movement of object is in line and direction of force.

OR

Work done is maximum, when displacement takes place along the direction of force.

Work done is given by the equation

W = F.S

<em> W = F. S cos Θ</em>

<em>When cos Θ = 0° ; cos 0 = 1</em>

3 0

3 years ago

Read 2 more answers

Write a question about how changing temperature affects gas

Lera25 [3.4K]

Answer:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

Explanation:

One possible question can be:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

The answer to this question is contained in Charle's law, which states that for a gas at constant pressure, the volume of the gas is proportional to its absolute temperature:

$V\propto T$

Or also written as

$\frac{V}{T}=const.$

By looking at this equation, we can find immediately the answer to our question: as the (absolute) temperature of the gas increases, the volume increases as well, by the same proportion.

3 0

3 years ago

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position

antiseptic1488 [7]

Answer:

The value of the distance is $\bf{14.52~cm}$ .

Explanation:

The velocity of a particle(v) executing SHM is

$v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)$

where, $\omega$ is the angular frequency, $A$ is the amplitude of the oscillation and $x$ is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., $x = 0$ .

The maximum velocity( $\bf{v_{m}}$ ) is

$v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Divide equation (1) by equation(2).

$\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Given, $v = 0.25 v_{m}$ and $A = 15~cm$ . Substitute these values in equation (3).

$&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm$

6 0

3 years ago